Gibbs Free Energy Calculation

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ΔG=ΔH-TΔSsystem and ΔSsystem=qp/T and isn't qp=ΔH? Which means ΔG=0 always which is obviously wrong.

But the derivation for Gibbs Free Energy is based on qsurrounding=-qsystem so shouldn't the term be equals to 0?
 

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  • #2
Ygggdrasil
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ΔG=ΔH-TΔSsystem and ΔSsystem=qp/T and isn't qp=ΔH? Which means ΔG=0 always which is obviously wrong.
ΔSsystem=qp/T is true only for reversible processes. What do you know about the ΔG of a reversible process?
 
  • #3
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ΔSsystem=qp/T is true only for reversible processes. What do you know about the ΔG of a reversible process?
It is 0. So for a reversible process qp=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks
 
  • #4
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It is 0. So for a reversible process qp=ΔH and so ΔG=0?

For an irreversible process how would the ΔS look like?

Thanks
Are you aware that ΔG and ΔS are functions only of the initial and final thermodynamic equilibrium states of a system, and are independent of path between the two states, whether reversible or irreversible? From what you have learned, do you know how to determine the change in S between two thermodynamic equilibrium states of a system? Are you familiar with the equation dG = -SdT + VdP, which applies to a differential change in temperature and pressure between two closely neighboring thermodynamic equilibrium states of a system?

I think your real question is "how do one determine ΔG and ΔS between two thermodynamic equilibrium states of a system."

Chet
 
  • #5
It is possible for ΔSsystem ≠ 0, even when qP = 0. Therefore, qP = 0, does not necessarily lead to ΔG = 0.

P. Radhakrishnamurty
 

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